CHAPTER 5 FIELD EFFECT TRANSISTORS(part a) (FETs)

Introduction of FET FETs (Field-Effect Transistors) are much like BJTs (Bipolar Junction Transistors). FETs sometimes called unipolar transistor operates only with one type charge carrier. The two main types of FETs are JFET (Junction Field Effect Transistor) MOSFET (Metal Oxide Semiconductor Field Effect Transistor) i. D-MOSFET i. D-MOSFET –– Depletion MOSFET ii. E-MOSFET ii. E-MOSFET –– Enhancement MOSFET The terms ‘Field Effect’ relates to the depletion region formed in the channel of a FET as a result of a voltage applied on one of its terminal(gate).

FETs vs BJTs Similarities: Amplifiers Switching devices Impedance matching circuitsDifferences: FETs are less sensitive to temperature variations and because of there construction they are more easily integrated on ICs. FETs are also generally more static sensitive (faster when turn on and off ) than BJTs. FETs are usually smaller than BJTs and particularly useful for IC chips. FETBJT Unipolar device – operate use only one type of charge carrier voltage controlled devices higher input impedance Bipolar device – operate use both electron & hole current controlled devices higher gains

FETs vs BJTs (a) Current-controlled and (b) voltage-controlled amplifiers. BJT – current controlled, I C is direct function of I B FET – voltage controlled, I D is a direct function of the voltage V GS applied to the input circuit.

JFET A representation of the basic structure of the two types of JFET. There are two types of JFET : n-channel and p-channel JFET Symbol

Basic Operation of a JFET  The channel width and the channel resistance can be controlled by varying the gate voltage – controlling the amount of drain current, I D.  The depletion region (white area) created by reverse bias.  Wider toward the drain-end of the channel – reverse-bias voltage between gate and drain is greater than voltage between gate and source.

JFET Analogy  JFET operation can be compared to a water spigot. The source The source of water pressure is the accumulation of electrons at the negative pole of the drain-source voltage. The drain The drain of water is the electron deficiency (or holes) at the positive pole of the applied voltage. The control The control of flow of water is the gate voltage that controls the width of the n- channel and, therefore, the flow of charges from source to drain.

8 Basic Operation of JFET

9 JFET Characteristics and Parameters, V GS = 0 Let’s first take a look at the effects with a V GS of 0V. I D increases proportionally with increases of V DD (V DS increases as V DD is increased). This is called the ohmic region (point A to B). In this area (ohmic region) the channel resistance is essentially constant because of the depletion region is not large enough to have sufficient effect  V DS and I D are related by Ohm’s law In JFET, I G = 0  an important characteristic for JFET

10 JFET Characteristics and Parameters, V GS = 0 At point B the I D cease to increase regardless of V DD increases. This called pinch-off voltage. As V DD increase from point B to point C, the reverse-bias voltage from gate to drain (V GD ) produces a depletion region large enough to offset the increase in V DS, thus keeping I D relatively constant.

11 JFET Characteristics and Parameters, V GS = 0 Continue increase in V DS above the pinch-off voltage produces an almost constant drain current  this drain current is I DSS (drain to source current with gate shorted). Breakdown occurs at point C when I D begins to increase very rapidly with any further increase in V DS. It can result irreversible damage to the device, so JFETs are always operated below breakdown and within the constant-current area (between points B and C on the graph)

12 JFET action for V GS = 0V

13 JFET Characteristics and Parameters, V GS < 0 As V GS is set to increasingly more negative by adjusting V GG. A family of drain characteristic curves is produced as shown in (b). Notice that I D decrease as the magnitude of V GS is increased to larger negative  causing the pinch-off is lowered as well (Boystead – lower in parabolic manner)

14 JFET Characteristics and Parameters, V GS < 0; V GS (off) As V GS becomes more negative: The JFET experiences pinch-off at a lower voltage (Vp). I D decreases (I D < I DSS ) even though V DS is increased. Eventually I D reaches 0A. V GS at this point is called V p or V GS(off) ( V GS (off) = V P ) Take note at Ohmic & Saturation Region Also note that at high levels of V DS the JFET reaches a breakdown situation. I D increases uncontrollably if V DS > V DSma x. FLOYD  V GS (off) = - V P ; reverse polarity

15 JFET Characteristics and Parameters, V GS < 0; V GS (off) For cutoff voltage (V G(off) ). The field (in white) grows such that it allows practically no current to flow through.

JFETTransfer Characteristics JFET Transfer Characteristics  The transfer characteristic of input-to-output is not as straightforward in a JFET as it is in a BJT.  In a BJT,  indicates the relationship between I B (input) and I C (output).  I C =  I B  In a JFET, the relationship of V GS (input) and I D (output) is a little more complicated:

JFET Transfer Curve This graph shows the value of I D for a given value of V GS. When V GS = 0; I D = I DSS When V GS = V GS (off) = V P ; I D = 0 mA

Plotting the JFET Transfer Curve Using I DSS and Vp (V GS(off) ) values found in a specification sheet, the transfer curve can be plotted according to these three steps: Solving for V GS = 0VI D = I DSS Step 1 Solving for V GS = V p (V GS(off) ) I D = 0A Step 2 Solving for V GS = 0V to V p Step 3

19 P-channel JFET operation  Same as n-channel JFET except required negative V DD and positive V GS.

20 EXAMPLE V GS (off)= -4V and I DSS = 12mA. Determine the minimum value of V DD required to put the device in constant-current region of operation when V GS = 0V.

21 JFET Transfer Characteristic

22 JFET Transfer Curve This graph shows the value of I D for a given value of V GS. When V GS = 0; I D = I DSS When V GS = V GS (off) = V P ; I D = 0 mA

23 EXAMPLE JFET with I DSS = 9 mA and V GS (off) = -8V (max). Determine I D for V GS = 0V, -1V and -4V. ANSWER: V GS = 0V, I D = 9mA V GS = -1V, I D = 6.89mA V GS = -4V, I D = 2.25mA